There exist many different equivalent formulations of Hensel's lemma in the literature. Usually the proof of their equivalence is difficult and uses deep knowledge of commutative algebra. I asked myself whether some similar-sounding statements have easier proofs. For example the following should have a rather easy proof, but I am struggling to find it.
Let $(R,\mathfrak{m},k)$ be a local ring. Then the following statements are equivalent:
$(i)$ For every $f(x) \in R[x]$ consider $\overline{f}(x)$ as its reduction mod $\mathfrak{m}$. If $\overline{f}(x) = (x-\overline{a}) \overline{h}(x)$ for some $\overline{h}(x)$ with $\text{gcd}(x-\overline{a},\overline{h}(x)) =1$ then one can find $h(x) \in R[x]$ such that $f(x) = (x-a)h(x)$ and $\overline{a},\overline{h}(x)$ are reductions of $a,h(x)$ mod $\mathfrak{m}$ respectively.
$(ii)$ For every $f(x) \in R[x]$ consider $\overline{f}(x)$ as its reduction mod $\mathfrak{m}$. If $\overline{f}(x) = \overline{g}(x) \overline{h}(x)$ for some $\overline{g}(x),\overline{h}(x)$ with $\text{gcd}(\overline{g}(x),\overline{h}(x)) =1$ then one can find $g(x),h(x) \in R[x]$ such that $f(x) = g(x)h(x)$ and $\overline{g}(x),\overline{h}(x)$ are reductions of $g(x),h(x)$ mod $\mathfrak{m}$ respectively.
Trivially $(ii) \Rightarrow (i)$, but for the other implication I did not find any readable proof unfortunately. $(i)$ is also clearly equivalent to same other rephrasings as for example to requiring that $\overline{h}(x)$ has $\overline{a}$ only as a simple root instead the $\text{gcd}$ condition.
Eisenbud covers both statements separately for complete rings as can be seen on page 206.
A proof of the equivalence above (including equivalence to many other statements) can be found in the lecture notes of Hochster on page 39 in http://www.math.lsa.umich.edu/~hochster/615W10/615.pdf . Nevertheless I am still hoping for an understandable and easy proof.
If you know it or a good reference for it, I would be very thankful. Thank you in advance.
